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The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

Short Answer

Expert verified
The minimum plate area needed is approximately 0.0136 m².

Step by step solution

01

Understand the Problem

We need to find the minimum area of the plates of a parallel-plate capacitor given specific electrical properties: dielectric constant, dielectric strength, capacitance, and maximum potential difference.
02

Use the Capacitance Formula

Recall the formula for the capacitance of a parallel-plate capacitor: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]where \(\varepsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \text{ F/m}\)), \(\varepsilon_r\) is the relative permittivity (3.60), \(A\) is the area, and \(d\) is the distance between the plates. Here, \(C = 1.25 \times 10^{-9} \text{ F}\).
03

Apply Dielectric Strength Condition

The dielectric strength provides the maximum electric field (\(E_{max}\)) the material can withstand, given by \(E_{max} = 1.60 \times 10^7 \text{ V/m}\). The potential difference \(V\) is related to the electric field by \(V = E_{max} \cdot d\). Solve for \(d\):\[ d = \frac{V}{E_{max}} = \frac{5500 \, \text{V}}{1.60 \times 10^7 \, \text{V/m}} \approx 3.44 \times 10^{-4} \, \text{m}. \]
04

Rearrange the Capacitance Formula for Area

We know \(C\), \(\varepsilon_0\), \(\varepsilon_r\), and \(d\), and we can solve for \(A\):\[ A = \frac{C \cdot d}{\varepsilon_0 \varepsilon_r} = \frac{1.25 \times 10^{-9} \, \text{F} \cdot 3.44 \times 10^{-4} \, \text{m}}{8.85 \times 10^{-12} \, \text{F/m} \cdot 3.60} \approx 1.36 \times 10^{-2} \, \text{m}^2. \]
05

Solution Conclusion

The calculation yields the minimum area of the capacitor plates necessary to meet the given requirements.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dielectric Strength
Dielectric strength is a critical property of insulating materials, especially ones used in capacitors. It refers to the maximum electric field that a material can withstand without breaking down or losing its insulating properties. In simple terms, it's the material's ability to resist an electric breakdown.
For a dielectric material, the dielectric strength is measured in volts per meter (V/m). A higher dielectric strength means that the material can endure a stronger electric field. This is essential in preventing breakdown in capacitors, ensuring the device functions correctly under high voltages.
In the context of capacitor design, knowing the dielectric strength helps determine how thin the material can be without risking failure. Choosing a material with a suitable dielectric strength ensures that the capacitor can operate safely at the intended voltage, as calculated in our exercise.
Parallel-Plate Capacitor
A parallel-plate capacitor is a straightforward type of capacitor that features two conducting plates separated by an insulating material called a dielectric. This setup is widely used due to its simplicity and effectiveness.
The capacitance of this kind of capacitor is determined by the area of the plates, the separation distance between the plates, and the properties of the dielectric in between. Larger plate areas and smaller separation distances increase capacitance, making for a more efficient energy storage device.
Parallel-plate capacitors are commonly used in electronic circuits and can be found in devices ranging from radios to laptops. Understanding the factors that influence their design, like the dielectric material and plate area, is key to optimizing their performance for various applications.
Capacitance Formula
The capacitance formula for a parallel-plate capacitor is: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]where:
  • \(C\) is the capacitance in farads (F).
  • \(\varepsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \) F/m).
  • \(\varepsilon_r\) is the relative permittivity or dielectric constant of the material.
  • \(A\) is the area of the plates.
  • \(d\) is the separation distance between the plates.
This formula calculates how much electric charge a capacitor can store. The greater the capacitance, the more charge it can hold for a given voltage.
By rearranging the formula, we can find different variables, such as the minimum area of the plates required for a specific capacitance, as done in the exercise. This is crucial for designing capacitors that meet precise electrical criteria.

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Most popular questions from this chapter

A parallel-plate capacitor has capacitance \(C_0\) = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m? (b) A dielectric with \(K = 2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m?

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

A 5.00\(\mu\)F parallel-plate capacitor is connected to a 12.0-V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, airfilled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 mm\(^2\), and the separation between the plates is 0.700 mm before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circuitry can detect a change in capacitance of 0.250 pF, how far must the key be depressed before the circuitry detects its depression?

A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 \(\mu\)C on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electricfield magnitude between the plates? (d) What is the surface charge density on each plate?

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